小编Ton*_*ony的帖子

未捕获的InvalidValueError:不是数组

我想知道是否有人可以指出我正确的方向,我正在使用Google地图试图为用户设置指定的邮政编码区域,如果我硬编码它的经度经度就像这样完美;

var triangleCoordsLS12 = [
        {lng: -1.558585,  lat: 53.796545}, 
        {lng: -1.558585,  lat: 53.796545},
        .....
];
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但是我试图从PHP数据库获取PHP和JSON这样的信息;

$.ajax({
          type:'POST',
          url:'test.php',
          success:function(data){
             var resultArray = JSON.parse(data);
               for (var i=0; i<resultArray.length; i++) {
                  var triangleCoordsLS12 = new google.maps.LatLng(resultArray[i].lat, resultArray[i].lng);

          if(location.uname == 'John Smith'){
            bermudaTriangleLS12 = new google.maps.Polygon({
                paths: triangleCoordsLS12,
                strokeColor: '#ff0000',
                strokeOpacity: 0.8,
                strokeWeight: 1,
                fillColor: '#ff0000',
                fillOpacity: 0.30
            });
           bermudaTriangleLS12.setMap(map);
        } else if(location.uname == 'Bruce Brassington'){
             bermudaTriangleLS12 = new google.maps.Polygon({
               paths: triangleCoordsLS12,
               strokeColor: '#FFcc00',
               strokeOpacity: 0.8,
               strokeWeight: 1, …
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javascript php mysql jquery google-maps

6
推荐指数
1
解决办法
8180
查看次数

org.xml.sax.SAXParseException根元素后面的文档中的标记必须格式正确

我知道有一些类似的线程,但没有一个对我有任何帮助,我正在尝试通过Java创建XML文件,但我在这条线上不断看到这个错误 doc = db.parse(is); ,有人可以请告诉我我要去哪里错了?错误:-

Severe:   [Fatal Error] :2:2: The markup in the document following the root element must be well-formed.
Severe:   org.xml.sax.SAXParseException; lineNumber: 2; columnNumber: 2; The markup in the document following the root element must be well-formed.
at com.sun.org.apache.xerces.internal.parsers.DOMParser.parse(DOMParser.java:257)
at com.sun.org.apache.xerces.internal.jaxp.DocumentBuilderImpl.parse(DocumentBuilderImpl.java:348)
at com.digivote.digivote.XmlServlet.doPost(XmlServlet.java:194)
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我的代码:

@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    response.setContentType("text/html");        
    String level = request.getParameter("level");
    String add = request.getParameter("add");
    String addCat = request.getParameter("addCat");
    String addQuestions = request.getParameter("addQuestion");
    String add_multiChoice = request.getParameter("add_multiChoice");

    try 
    { …
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java xml

3
推荐指数
1
解决办法
6288
查看次数

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